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Lemma 11.4.3. Let V be a k vector space. Let K be a central k-algebra which is a skew field. Let W \subset V \otimes _ k K be a two-sided K-sub vector space. Then W is generated as a left K-vector space by W \cap (V \otimes 1).

Proof. Let V' \subset V be the k-sub vector space generated by v \in V such that v \otimes 1 \in W. Then V' \otimes _ k K \subset W and we have

W/(V' \otimes _ k K) \subset (V/V') \otimes _ k K.

If \overline{v} \in V/V' is a nonzero vector such that \overline{v} \otimes 1 is contained in W/(V' \otimes _ k K), then we see that v \otimes 1 \in W where v \in V lifts \overline{v}. This contradicts our construction of V'. Hence we may replace V by V/V' and W by W/(V' \otimes _ k K) and it suffices to prove that W \cap (V \otimes 1) is nonzero if W is nonzero.

To see this let w \in W be a nonzero element which can be written as w = \sum _{i = 1, \ldots , n} v_ i \otimes k_ i with n minimal. We may right multiply with k_1^{-1} and assume that k_1 = 1. If n = 1, then we win because v_1 \otimes 1 \in W. If n > 1, then we see that for any c \in K

c w - w c = \sum \nolimits _{i = 2, \ldots , n} v_ i \otimes (c k_ i - k_ i c) \in W

and hence c k_ i - k_ i c = 0 by minimality of n. This implies that k_ i is in the center of K which is k by assumption. Hence w = (v_1 + \sum k_ i v_ i) \otimes 1 contradicting the minimality of n. \square


Comments (2)

Comment #4637 by Juan Sebastian Gaitan on

I think in the third line of the proof (line 224 of the .tex file), should be changed to . The lack of parenthesis is confusing.

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